Question

The equation of the tangent at \( P(-4, -4) \) on the curve \[ x^2 = -4y \] is

• A. \( 2x + y + 4 = 0 \)
• B. \( 2x - y + 4 = 0 \)
• C. \( 2x + y - 4 = 0 \)
• D. \( 3x - y + 8 = 0 \)

Hint:

To find the equation of a tangent to a curve, use the derivative to find the slope and apply the point-slope form of the line equation.

Answer 1

The Correct Option is B

Step 1: Finding the equation of the tangent.
For the curve \( x^2 = -4y \), the derivative is: \[ \frac{d}{dx}(x^2) = \frac{d}{dx}(-4y) \quad \Rightarrow \quad 2x = -4 \frac{dy}{dx} \] Thus: \[ \frac{dy}{dx} = \frac{-x}{2} \] At the point \( P(-4, -4) \), the slope of the tangent is: \[ \frac{dy}{dx} = \frac{-(-4)}{2} = 2 \]
Step 2: Equation of the tangent.
Using the point-slope form of the equation of a line, we have: \[ y - (-4) = 2(x - (-4)) \quad \Rightarrow \quad y + 4 = 2(x + 4) \] Simplifying, we get: \[ 2x - y + 4 = 0 \]
Step 3: Conclusion.
Thus, the equation of the tangent is \( 2x - y + 4 = 0 \), which makes option (B) the correct answer.